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29 points

·
10th Jan 2017

I have a couple cheat sheets from my machine learning course last semester:

https://www.sharelatex.com/project/5858a8dc4fa3ca7871c17f89

https://www.sharelatex.com/project/580ccb003cbcc62e38eb21ca

They are slightly incomplete b/c I only wrote as much as I could in the 2-3 days before the exams, but there's quite a bit of good stuff there.

28 points

·
7th Apr 2018

Good question.

Differentiation (taking derivatives) only preserves **identical** equality.

The "=" symbol can actually have two meanings.

1) Identity. In this case, f(x) = g(x) means "the function f is equal to the function g for all choices of x. No matter what x I choose, f and g give the same output". An example of this are trig identities, (cos(x))^2 + (sin(x))^2 = 1 **for all x**. If you have an equality of this type, then you CAN differentiate and the two things will remain equal.

2) Conditional equality. This means that f(x) = g(x) for **some choices of x, but not all.** Differentiation does NOT preserve this equality.

https://www.desmos.com/calculator/yo3jflsj5y

Here, you can see the graphs of x^2 , x^3 , and their tangents at the point (1,1).

As you can see, x^2 =x^3 at (1,1), but their slopes (the slope of the lines) are not equal. This is because x^2 is not identically equal to x^3.

Edit: **I should also add, differentiation preserves functions that would be identically equal if you shift them.**

For example, Let f(x) = g(x) +3 for all x. g'(x) will be identically equal to f'(x). **However, this isn't useful for your exact question because if f(x) = g(x) + c (for all x) where c is a real number other than zero, then the two functions can never be equal**

27 points

·
27th Oct 2019

William Dunham has a great book,Journey through Genius: The Great Theorems of Mathematics, about this.

18 points

·
10th Sep 2019

I highly recommend reading "Mathematical Proofs: A Transition to Advanced Mathematics" by Gary Chartrand et. al. It helped me get a better understanding of how to write a proof as well as organize my own thoughts.

Here's the Amazon link: Mathematical Proofs: https://www.amazon.com/dp/0321797094/ref=cm_sw_r_cp_apa_i_V1UDDb4JBGWFX

17 points

·
1st May 2021

Based from the comments below I believe I can make a safe assumption that most people are given the impression from Brilliant.org that this website is a great way to learn mathematics on a deep level.

I am not one of those people. Based from the website's advertisement it's more of a fun learning course than a serious in-depth learning course of any particular topic. It's mainly used to challenge you to think differently in specially-designed puzzles.

17 points

·
18th Jul 2021

I can't think of a single book that captures this so cleanly. A few that have been useful:

- Kevin Houston's How to think like a mathematician?
- Lara Alcock's How to study as a Mathematics major? or any of her books on analysis.
- Polya's books on discovery.
- In general studying books that are Inquiry Based Learning focused. Example Burns's books on analysis/group theory/number theory or e.g. David Farmer's short books on group theory/knot theory. A lot of these books present theorems in a sequence of chunks which are intended to capture the process of how one would think of proofs for them.
- Oh I also like Devlin's book that someone else recommended.

16 points

·
19th Dec 2020

The texts by Stewart and Larson are the two most common introductions to calculus, and both are fantastic. Calculus is such a powerful tool that you don't need to worry about tailoring it to one field. Just as addition and multiplication will be used by economists and physicists alike, so will integration and differentiation.

14 points

·
13th May 2018

There are definitely ways to visualize algebraic concepts and many algebraic concepts crop up in geometry. Unfortunately, many books and classes won't emphasize visual intuition. So algebra may be harder for you. In some ways, you get over it even if it isn't your cup of tea, but there are also resources for transfering visual/geometric intuition onto algebraic concepts.

After reading it for myself, I recommend the books visual group theory by Nathan Carter, and algebra, concrete and abstract by Frederick Goodman. The first focuses a lot on visual intuition for group theory, but a lot concepts in group theory generalize to abstract algebra in general. The second book is a more traditional book, less focus on visual intuition, uses symmetry of geometrical objects and linear algebra for many of the examples.

13 points

·
25th Sep 2012

Those are all quadratic equations. I'd look them up at Khan Academy: http://www.khanacademy.org/math/algebra

There are several ways of solving them, but if you're not already quite familiar with them, attempting to teach them in a post here won't work too well.

See the videos on solving them by factoring, by using the quadratic formula, and completing the square.

13 points

·
11th Oct 2018

It's a **sufficient** condition, i.e. if the condition is true then f(x) is definitely greater than g(x). It is not a **necessary** condition, as you can find plenty of examples where f(x) > g(x) but the condition doesn't hold, such as this example.

13 points

·
20th Jun 2011

I don't know if you're aware of the Khan Academy, but Khan's videos are an immense help when you're puzzling out new math topics. I highly recommend watching some of his vids.

13 points

·
17th Apr 2018

Try picking up a book. I recommend this one. You can also use Rudin but it will be more difficult.

If you are using notes and online research, it may be that the exercises you've been working on are coming from many different areas and aren't really focused on one topic in particular. This may be the reason that every problem seems to require a new trick.

While it's certainly not the best or broadest advice, I've always found that, whenever a problem starts to get excessively complicated, the mean value theorem always seems to be the *why-didn't-I-think-of-that* trick that solves it.

12 points

·
13th Jul 2017

Obviously not a book, but...Calculus revisited, the old 1970's b&w MIT lecture series by Herbert Gross, is the best lecture series I've ever found for gaining a solid intuitive understanding of calculus. If you're like me, the preface will even get you pumped. There are more recent MIT openCoursWare calculus videos, but, IMO, none of them compare to this oldie but goodie. He's so good at explaining things that you can even just listen to the audio alone and most of the time you'll get the idea.

IDK why all the older material seems to be the best, but another great resource is Calculus Made Easy by Silvanus P. Thompson. 1910. It practically starts out at the kindergarten level, but advances quickly. The analogies and intuitive explanations are spot on, and there are worked examples provided.

Both are old-as-shit, free, and IMO at least twice as good as anything else out there.

12 points

·
29th Jun 2018

I answered your question on r/cheatatmathhomework

But since you want a general formula, here's something:

Let n be the number of good cards you need to get before n bad cards.

Let GT be the number of total good cards

Let BT be the number of total bad cards

Let TT be the total number of cards.

Let k be the number of moves before the game terminates.

The minimum number of moves to win is n, and the maximum number of moves to win is 2n-1.

There will be **(k-1) choose (k-n)** winning variations of length k.

A variation of length k has probability of **[(GT!)(TT-k)!(BT!)] / [(TT!)(GT-n)!(BT+n-k)!]**

So in the end, your probability of winning is:

**sum from k = n to k = 2n-1 of:**

**[(k-1) choose (k-n)]*[(GT!)(TT-k)!(BT!)] / [(TT!)(GT-n)!(BT+n-k)!]**

__________________________

Here's the formula in action when you need to flip 3 good cards to win, you have 4 good cards, and 5 bad cards. https://www.desmos.com/calculator/picz8zb7zp

_____________________________

Explanation (in case you wanna know):

There's (k-1) choose (k-n) variations of length k because we know for you to win, you must terminate on a good flip. That means there's k-1 positions which can have either good or bad flips, but we know the number of bad flips is exactly k-n, so we have k-n bad flips to place over k-1 positions, hence k-1 choose k-n permutations.

For the second term, we must have the biggest n numbers of GT!, and hence we have GT!/(GT-n)!, this ensures the variation is winning (we've drawn the first n good cards). We have (TT-k)!/(TT!) which ensures we've made n total flips. The term (BT!)/(BT+n-k)! counts the number of bad flips.

12 points

·
29th Apr 2018

A way you can see the relationship graphically is to go to the desmos online graphing calculator

https://www.desmos.com/calculator

1) Type in a^x for one function

2) Type d/dx a^x for another function

3) Set the slider for a to be between something like 1.8 and 10, and set it to step in increments of .01

When you click play you will see that the graph of the derivative moves back and forth from one side of its associated exponential graph to the other side. There is one brief moment where the derivative graph passes right on top of the associated exponential graph and they are the same. This happens at a = e

12 points

·
12th May 2019

Ordinary Differential Equations from the Dover Books on Mathematics series. I Just took my final for Diff Eq a few days ago and the book was miles better than the one my school suggested and is the best written math textbook I have encountered during my math minor. My Diff Eq course only covered about the first 40% of the book so there's still a TON of info that you can learn or reference later. It is currently $14 USD on amazon and my copy is almost 3" thick so it really is a great deal. A lot of the reviewers are engineering and science students that said the book helped them learn the subject and pass their classes no problem. Highly Highly recommend. **ISBN-10:** 9780486649405

​

https://www.amazon.com/gp/product/0486649407/ref=ppx_yo_dt_b_asin_title_o08_s00?ie=UTF8&psc=1

12 points

·
22nd Jun 2018

Read and work through <em>Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach</em> by Hubbard and Hubbard - can be bought pretty cheaply used. It's a book that starts at introductory linear algebra and infinite series and ends by covering vector calculus with a treatment of differential manifolds.

Greatest introductory university maths book I've ever taught from or read. Has enough material to take you through at least 3 academic semesters of work if you go through it slowly and will put you in a great position to understand a more rigorous analytical text like Walter Rudin's *Principles of Mathematical Analysis* upon completion.

11 points

·
23rd Jul 2011

Rather than plotting both of them, it is a little clearer if you plot their difference: http://www.wolframalpha.com/input/?i=-2%2F3+log%28cos%283x%2F2%29%29++%2B+1%2F3+log%281+%2B+cos%283x%29%29

The difference (barring the problems with log of a negative number) is constant, and thus their derivative is the same, ie they are antiderivatives of the same thing.

11 points

·
5th May 2012

First, congratulations on deciding to further your knowledge of basic math, you might decide to check out Khan Academy as you get to other subjects that you might have previously had trouble with. The answer to your question is that you have a "remainder" of 4. That means that 12/8 is actually equal to 1 + (4/8), or 1 plus 4 over the original denominator. That's why the calculator gives you .5 , you don't just put the remainder after the decimal point, you put it over what was originally on bottom to find the decimal.

11 points

·
20th Mar 2011

Try taking a look at the graph of x^x, then the graph of x^n (for different choices of n). When x is positive, then for whatever n you choose, the graph of x^n will be either increasing or decreasing (or both if n = 0). So, the behavior of the graph is determined by the sign of n. However, if you look at x^x on the interval [0,1], you'll see that it *decreases* at first, even though the values of both pieces of x^x are positive. So, the slope of the tangent line will be negative at those spots, but x^x is always going to be positive when x is positive. The power rule cannot work, then.

Edit: Some rephrasing for clarity.

11 points

·
23rd Jun 2018

If you want a general method:

https://brilliant.org/wiki/linear-diophantine-equations-one-equation/

diophantine equations are a standard thing in number theory.

But, in this case, it might actually be quicker to just do it brute force.

5x+8y=(x,y)=38

[I don't know what you mean by "probability tree." So I'm probably just doing what you did.]

There are only six possibilities for y and three of those are very obvious and the other three are simple. So you can go through those very quick.

**0.**y=/=0 because 38 ends in "8" not 0 or 5.

**1.**y=1 implies x=6

*This is a solution.*

y=2 implies

5x+16=38

x=(38-16)/5=4.4

**2.** y=/=2

y=3

5x+24=38

x=(38-24)/5=2.8

**3.**y=/=3

y=4

implies

x=(38-32)/5=1/2

**4.**y=/=4

The only other possibility is all fives but that can't be correct.

So there's only one solution.

11 points

·
24th Nov 2019

Buy this book first.

Learn Python the Hard Way: A Very Simple Introduction to the Terrifyingly Beautiful World of Computers and Code (3rd Edition) (Zed Shaw's Hard Way Series) https://www.amazon.com/dp/0321884914/ref=cm_sw_r_cp_apa_i_WgE2Db53ZBZ58

Then set aside 1-3 hours a day to do the lessons, read the book and/or look up additional material on the internet.

The first chapter or the introduction tells you to try and memorize basic powershell commands, its useful but don't spend too much time trying to get them all down.

10 points

·
5th Jul 2021

When it comes to tutoring, it does not really matter what you use to spread knowledge - textbook, Brilliant.org, whatever. The only thing that matters is You. A great tutor who is excited to teach their subject is a gem in education, regardless of what they use, so keep that in mind. Good luck in your tutoring journey!

9 points

·
14th Aug 2013

There is a reason your teacher has used ten days of class to talk about the methods and techniques for finding derivatives. There are lots of rules to learn, and you need to practice them. It isn't something that can be taught and learned in a paragraph or two on an Internet forum.

If you don't understand what has been taught in class, read your textbook, or go get a calculus study guide, or try Khan Academy.

9 points

·
3rd Feb 2012

Try the excellent 'Khan Academy'. It has many youtube videos of various math concepts explained.

There is also a 'fun' workplan geared towards kids:

http://www.khanacademy.org/exercisedashboard

Please let us know how you find it. It's got many recommendations on here.

9 points

·
9th Jun 2018

I would recommend putting them in an anki deck. It's a spaced repetition program that is tremendously useful for memorization. There is an accompanying phone app (though I think the iOS version might not be free?), so you can easily go over all the formulas you want to memorize as they come up every day. Anki uses an algorithm to decide how often you need to train a certain piece of knowledge based on how well you know it already.

On top of that, I also recommend actually deriving them yourself at least once just so you're sure you understand where they come from. It's also very important to actually do exercises where you need those identities. Recognizing when you need to use what is part of the difficulty and that can only be taught by doing it.

8 points

·
9th Sep 2011

Here is a link to the practice exercises portion of the site.

Here's a video explaining how the exercises work.

8 points

·
21st Jul 2019

Most mathematical documents are made using something called LateX. There are a number of different ways to do it but I think the most popular is through a website named Overleaf. It has a lot of tutorials and uses a cloud like Google docs.

8 points

·
21st Feb 2017

Since you seem to need a bit more of a push, I will answer the question I wrote in my other reply to this post.

Watch what happens when I input x = -3 into this relation:

(-3 + 3)^2 + (y - 1)^2 = 4

(0)^2 + (y - 1)^2 = 4

(y - 1)^2 = 4

Now what number, or numbers, can be squared so that you get 4?

There are *two* of them: 2, and - 2.

This means that we have two possibilities:

(y - 1) = 2

or

(y - 1) = -2

In the first case, we find that y = 3, and in the second case we find that y = -1.

So when we *input* x = -3, we get **two** *outputs*: y = 3, and y = -1.

Since we found an input that is related to **more than one** output, this relation is **NOT** a function.

Alternatively, if you graph this relation, you get a very familiar shape; one that I'm sure you've seen in previous math courses.

Notice how when x = -3, there are **two** points on the graph. This means that the input x = -3 is related to **two** outputs.

And therefore, it is not a function.

8 points

·
23rd Apr 2018

In my opinion, sounds like all you need to do is make sure you work hard once you begin the course. Currently I'm a junior math undergrad, but I had a late start to my math education. I never was "good" at math in grade school, always in the average math class. I didn't take one senior year in high school. I actually failed college algebra TWICE my freshman year of college, because I was skipping class. After that, i decided that I actually love math and just started to work hard, that was all, i just put in a lot of effort.

I went to a different school and just went to trig because of test scores, got an A. Then went back to my university, and have gotten A's in Calc 1-3, Linear Algebra, and Complex Analsysis. Literally only because I just worked really hard. I found a study partner and we did many hw assignements together and studied for exams together which helped too.

So it already sounds like you are 5 steps ahead of where I was in your position. I think you shouldn't worry as long as you do each homework, study for exams, and think about what you are doing in class conceptually. Talking it out with someone is a great way to retain that info, and explaining it to someone is even better. I got a job at the tutoring center at my university so it would force me to have to explain the math to other students, maybe that is an option at your school you could look into.

edit: spelling

edit 2: i guess this wasn't directly answering your question of resources that you could use to help you preapre. I guess in my mind you sound quite prepared, and I am just trying to ease your worries about univeristy calc, its not so bad! It's actually great fun :) Check out brilliant.org for some fun exercises

8 points

·
2nd Jun 2021

Still good? Yeah, they're a foundational text in mathematics. There are some really beautiful illustrated versions of *Elements* out there.

7 points

·
30th Jun 2012

Khan academy explains even the simplest things quite thoroughly. I suggest you check it out http://www.khanacademy.org/math/algebra If you register (free), it can track your progress and help you choose what to watch next. Good luck!

7 points

·
10th Jun 2020

At the very most general level, programming would really consist of

- Breaking down a problem into smaller pieces very typically some sequence of "actions"
- These sequence of actions almost always breaks down into: a) selection - pick the next action or item given the result of some other action. b) doing an action on some item to generate some result (it could be TRUE/FALSE, or some mathematical calculation)

The reasoning skill behind programming is therefore not at all different from almost all mathematical problems. Math pretty much does the same thing. The main difference is, of course, repeated action is done very quickly by a computer whereas doing things by hand is tedious. Most of the reasoning skill required is based on "binary logic" which is a fancy term to say breaking down big problems into TRUE/FALSE or YES/NO series of questions/actions.

​

If you feel you are an absolute beginner, try something like SCRATCH. If this feels too simple for your level, PYTHON is a nice next step. If you really want to dig deeper into programming C and C++ is also good but has a steeper learning curve. There are quite a number of free Python and C online tools/environments. There are other tools that are more specialized, for example focused on math, like MATLAB, or simulation. Others might be focused on AI/neural networks. Generally though a good foundation in programming basics would be the recommended path (might take a few months to a year) before embarking on a deep dive.

7 points

·
31st Oct 2017

You are not stupid!!

Fortunately, we live in an age where there are free tools to help you understand.

Check out this graph, and play with the sliders!!

(Note that you can also edit the values for a, b, and c , not just move the sliders).

To help you with your intuition for Bx - try setting a = 1 and c = 0. Then move the slider for b. What do you notice about the way the curve behaves? Now test your intuition by doing the same thing with different values of a and c.

You are asking the most important question to make progress in mathematics, which is "what is the intuition for this?" You are thinking like a mathematician :).

7 points

·
12th Jun 2018

Most textbooks I know of on PDEs are quite dull and old-fashioned.

You could start with Paul's Online Math Notes

http://tutorial.math.lamar.edu/Classes/DE/IntroPDE.aspx

Or Scholarpedia

http://www.scholarpedia.org/article/Partial_differential_equation

7 points

·
7th May 2011

<strong>This</strong> might be what you're looking for. Explore the site. It has lectures in a great range of subjects from the basics all the way till college level. The site also has a testing portion with earn-able medals and achievements.

I would say you should force yourself to work on math for a few hours a day until you reach your desired goal. If you need motivation, /r/GetMotivated's got your back!

7 points

·
13th Jun 2021

G.H. Hardy once wrote a book called A Mathematician's Apology, in which he apologized to the British public for wasting their tax money on studying "useless" mathematics, simply because he thought it was beautiful. That "useless" mathematics has since become the foundation of all internet encryption methods.

On the other hand, Edward Lorenz was a mathematician until he was recruited to be a weather forecaster for the US Army Air Forces during WW2. After the war, even though he was only a few months away from his PhD in mathematics, he switched to meteorology because he felt like there were fundamental questions in the field that he could contribute to solving. He went on to discover what we call "the Butterfly Effect," the foundation of Chaos Theory.

Bottom line: You never know. It's useless to pick a field simply because it's "cooler" or "more useful," because you never know what's going to come your way, and you never know what other people are going to do when your work passes on to them. Go with what interests *you* the most, and see where that takes you.

7 points

·
11th Jun 2021

I used to be a dealer too! I ended up dealing blackjack, roulette, craps, bacc, Pai Gow, and many poker variants. I recommend that you read the book Secrets of Mental Math or watch the dvd. To practice, look at getting an app called Anki and make flashcards for yourself. Good luck and have fun!

6 points

·
3rd Jun 2017

This is the Hamming weight of the number, if you're in binary. For something like ternary, then the Hamming weight would be the number of digits that aren't 0 in the ternary expansion.

It's calculated pretty much by just counting the number of 1s in the binary expansion of the number, but the implementation mostly varies at the CPU level. Some processors have an instruction that can efficiently calculate it (which is known as popcount), while others will split the number into a bunch of parallel jobs and count it that way. This stack overflow link is a good explanation of different implementation for calculating it.

6 points

·
9th May 2013

The answer lies in etymology.

From an etymological dictionary.... >-teen combining form meaning "ten more than," from Old English -tene, -tiene, from Proto-Germanic *tekhuniz (cf. Old Saxon -tein, Dutch -tien.

>The combining form of ordinal numbers, -teenth, developed from Old English -teoða, -teoðe (West Saxon), teogoða (Anglian) "tenth."

Eleven, is derived from old English and predates -teen; "endleofan" means one [left]over(ten)

Similarly, twelve predates the use of -teen; "twelf" means two left (over ten).

The use of the suffix -teen and calling 13 "thirteen" (in the 15th century) simply came after the numbers 11 and 12.

6 points

·
25th Mar 2012

You need to know two triangles - 30/60/90 and 45/45/90.

Once you know the relationships between the legs and hypotenuse for each triangle, sohcohtoa will take you the rest of the way.

Kahn has a decent video on the two triangles.

6 points

·
5th Jun 2018

Variables will be used for number of orange drinks and number of colas.

j = number of juice

c = number of cola

Then write expressions for cost of juices and cost of colas. This will be in dollars, so use decimals to represent cents.

0.60 * j = cost of j juices

0.70 * c = cost of c colas

Total cost is known to be $10.00. So add the above expressions and say the sum is 10.

0.60j + 0.70c = 10

You could graph this on a coordinate plane where horizontal axis is j values and vertical axis is c values. A tool like desmos.com does this quickly.

https://www.desmos.com/calculator/vxzgdzyvck

If you it by hand you might rewrite this in slope intercept form.

0.7c = 10 – 0.6j

c = 10/0.7 – 0.6j/0.7

c ≈ 14.28 – (6/7)j

c ≈ (–6/7)j + 14.28

The intercepts are not whole numbers, but if you look along the graph, see if there are any points **where the line crosses integer grid lines**.

The problem statement suggests there are two such "lattice points." Those are the two solutions.

We disregard the other positions on the line because fractional number of drinks doesnt make sense. Also disregard places where line is in regions where c or j is a negative number.

Last step is to interpret the answer. (XX, YY) as an answer point means XX juices and YY colas. Check the answer by seeing if that many of each type would indeed cost $10.

*There are other lattice points near but not on the line. Those would be numbers of drinks that are close to costing $10 but actually cost a little more than $10 (if above the line) or a little less than $10 (if under the line).*

This is not really a "systems of equations" problem (which uses more than one equation), but in a sense you are looking for intersections of the diagonal line with horizontal and vertical grid lines.

6 points

·
16th Jan 2015

I'd counter this with a strong recommendation for Keith Devlin's "Introduction to Mathematical Thinking" class. He's a prof at Stanford who has been training up-and-coming mathematicians for years.

One of the wonderful things about math is that it's really not about memorizing things, and his class helps you develop the right ways of thinking to be successful in math.

The course starts 2/14: https://www.coursera.org/course/maththink

6 points

·
28th Jun 2018

Have you looked at all into https://brilliant.org yet? Not only are some of the problems challenging in terms of some of the mathematics involved (although you can search by subject), but many will require a lot of thought about what types of mathematics are required (or not) for a problem.

Is this what you're looking for when you say you're looking for challenging math problems?

6 points

·
21st Sep 2020

I always enjoyed connecting the values of the trig functions to the unit circle. Each of the functions has a geometric interpretation which helps with gaining intuition. Check out this link for example: https://www.geogebra.org/m/keqhdkaj

6 points

·
5th Apr 2011

Try Khan Academy. Dozens of videos on calc topics. Very intuitive and helpful presentation. There are exercises on the site too but I've never done them so I can't judge their usefulness.

The other thing to note is that nothing will "teach" you Calculus except practice. Lots of practice. Do as many of the practice problems in your book as your brain will allow. Then do two more. Do them even if you can't check the answers in the back (but do the ones w/ answers first). I don't mention this to be dickish or pedantic, but because it is the truth.

6 points

·
20th Feb 2011

Instructions:

Head to www.khanacademy.org.

Click on Login in the upper right and go through with the log in. You will need an account with google or facebook.

Click on Do Exercises at the top left, just under the "xx,xxx,xxx lesson delivered" line under the logo.

You should now be presented with a tech tree that starts with Addition.

6 points

·
12th Jun 2018

The book I used for two semesters was Introduction to Partial Differential Equations by Olver. I found this to be a good book, not amazing but good. Quite a few typos but there is an errata on either Springer's or Olver's site (can't remember which) along with a PDF of selected solutions. I used it for a class but I could see self study from it being doable. I have been told by some Professors and friends that Applied Partial Differential Equations with Fourier Series and Boundary Value Problems by Haberman is a staple. I have not read it though so I can not personally speak on its quality. I've attached Amazon links for both but, with enough digging I'm sure you could find a PDF of each online.

Olver - https://www.amazon.com/Introduction-Differential-Equations-Undergraduate-Mathematics/dp/3319020986

Haberman - https://www.amazon.com/Applied-Partial-Differential-Equations-Boundary/dp/0130652431

6 points

·
17th Oct 2016

If you work on geometry problems long enough, you'll realize that there are four 45 degree angles in 180 degrees. 180 - 45 = 135. So 135/45 = 3.

The book *Surely You're Joking, Mr. Feynman!* has a chapter on random number tricks like these (it's not comprehensive, but he does explain some of them). Someone asked him for the cube root of 1729 and he knew that 12^3 = 1728 (number of cubic inches in a cubic foot), so he just wrote 12 and a little bit.

5 points

·
7th Jun 2015

Coursera is a free MOOC provider that has a number of Mathematics Courses. Their Calculus One course by Jim Fowler from The Ohio State University has open enrollment and is self paced.

5 points

·
17th Jun 2011

it is worth noting that the real roots don't have any intrinsic geometric meaning either - they are just those values of x that make x^2 - 2x = 0 in exactly the same way that the complex roots of x^2 + 2 are the values of x that make x^2 + 2 = 0. the fact that a real-valued function of real numbers can be represented in the 2d plane has nothing to do with the function itself, and the "x axis" is simply the set of all points where y = 0, so the fact that the parabola crosses the axis at the roots is a tautology.

the only difficulty with extending this to the complex numbers is that a complex-valued function of complex numbers takes four dimensions to represent (`f(a+bi) = u+vi`

), but if we did the complex roots would be those places at which the hypersurface (a + bi)^2 + 2 crossed the u = v = 0 hyperplane. check out the wolfram alpha plot of (a + bi)^2 + 2 and look for those points where both the real and imaginary part are simultaneously zero.

5 points

·
25th Jun 2012

It does indeed. http://www.khanacademy.org/exercisedashboard

I've been working my way through it. Its pretty neat actually. Each activity has the list of associated videos. You're required to get a decent sized run of problems correct in a row to be declared competent in any topic. It periodically nags you to complete review problems in topics you previously completed. If you get stuck it can give you tips although it penalises you if you ask for them.

5 points

·
23rd Feb 2012

That is just an application of exponent rules . That video and the next few in the series should give you a good grasp on how they work.

5 points

·
21st Feb 2012

It would be much easier to point you in the right direction if you gave us a couple examples of the problems you're having trouble with.

Regardless, Khan Academy is a fantastic site that covers all of the math you're likely to see throughout high school.

5 points

·
2nd Jul 2018

They differ only by a constant, which means that your two results are both antiderivatives of (x-1)^(2).

This is similar to how x^2 and x^2 + 1 are both antiderivatives of 2x.

5 points

·
20th Nov 2011

Here's a great video from the Khan Academy about the intuition behind logarithmic scales.

Here's another one that deals with more of the real math of it.

5 points

·
31st Jan 2021

I second the recommendation of Overleaf. If you need to be able to compile your code offline, though, then Overleaf wouldn't fit your needs. I should also add that /r/LaTeX is another good resource here on reddit for all things LaTeX-related.

Good luck!

5 points

·
18th Nov 2018

Oh man, your method is absolutely beautiful. That substitution just makes everything fall into place, I didn't even need anything else after that since it simplified the problem massively.

I do have 1 question though, why do you put your dx's before the integrand?

EDIT:I went back to my proof and it turns out if you write out arctanh(t)/t as a taylor series around t=0 it actually immediately turns into the end of your proof, no polylogarithm needed.

5 points

·
3rd May 2015

Thanks for your help, much appreciated.

I have attempted to elaborate, however I am struggling with the surjection proof.

https://www.overleaf.com/read/scmwzqwncmjb

EDIT: just completed a few extra edits

5 points

·
1st Apr 2015

Do you know how to graph inequalities?

Let's start with the first part.

"It costs 15 dollars/week to make a rocker and 25 dollars/week to make a glider the manufacturer has 4500 dollars to spend"

So that means that some combination of 15 dollar rockers (x) and 25 dollar gliders (Y) can't be more than 4500 dollars or

15x+25y≤4500

"The facility can only hold 200 pieces of furniture."

So that means the number of rockers (x) and gliders (y) can't be more than 200 or

x+y≤200

When we graph 15x+25y≤4500 and x+y≤200 on Desmos it looks something like this https://www.desmos.com/calculator/nfxeshg49s

"The profit on each rocker is 60 dollars and the profit on each glider is 65 dollars."

That means our objective function (our business goal) is 60x+65y

When we use the objective function we look at the corners where all the shading overlaps.

These points are (0,180),(50,150) and (200,0)

Then, we plug these points into our objective function which is 60x+65y to find the maximum profit.

60(0)+65(180)= 11,700 60(50)+65(150)= 12,750 60(200)+65(0)= 12,000

So it looks like 50 rockers (x) and 150 gliders (y) will get you the most profit

Sorry for the weird formatting. Let me know if you have any more questions!

5 points

·
20th Feb 2011

How would you graph y = 1 - x?

Well, you know the graph goes through y = 1 and has a slope of -1. You can visualize this by looking at the function. Why can you visualize this? You have worked with this idea before. You may have started learning this by choosing x's and finding y's.

What I am getting to is: you are laying the groundwork to make these polar equations easier to visualize/graph by doing many simple ones by hand. Intuition will come the more you work the problems.

Also, if you're just doing it because you have to and you don't really care, Wolfram Alpha.

5 points

·
18th Jul 2011

Oh, man. Most of my recommendations would be this site, but for Algebra... It's just perfect.

Apart from the videos, there are about 115 exercises covering arithmetic and algebra.

Look through the playlists; you'll want to use others than the main "Algebra" one, especially the worked examples ones and the more advanced Developmental math videos.

5 points

·
18th Jul 2011

Khan Academy has exercises as well.

There are many many exercises you can do from and they really do start from the most basic of basics. Start from the very top and do a few questions. If you think its too easy, skip the next few exercises until you find something that you do not know but you feel is within your reach. Perfect that exercise, rinse and repeat. And of course, practice a lot.

If you need any specific help, you can always post again in /r/learnmath.

5 points

·
1st Oct 2016

Art and Craft of Problem Solving by Zeitz

Just search amazon for Art of Problem Solving and you'll get a tonne of resources

5 points

·
17th Sep 2020

Certainly no expert, but if you have a little calculus, and some background in statistics, maybe machine learning could be fruitful to look in to? That is if you don't use that sort of thing already. Certainly if you already have database/data extraction experience your time with ML is going to be much easier.

edit: As far as books, Make Your Own Neural Network by Tariq Rashid had utility to me.

4 points

·
25th Jul 2013

Understand what? If you're looking for resources, go take a look at Khan Academy or PatrickJMT. There are plenty of videos and practice problems to help you out.

If you have a more specific question or problem set for us to look at, post that instead. Try not to be so vague in your title and include some sort of description to help us help you. Good luck!

4 points

·
13th Nov 2012

Unfortunately, there are no shortcuts to learning mathematics.

"There is no royal road to geometry."

- Reply given when the ruler Ptolemy I Soter asked Euclid if there was a shorter road to learning geometry than through Euclid's Elements.

4 points

·
23rd Nov 2015

Two suggestions:

1) Keith Devlin's Coursera course Introduction to Mathematical Thinking

2) Learning to Reason by Nancy Rodgers. Excellent text that assumes nothing more than high school math.

4 points

·
24th Jun 2011

Disclaimer: my first time working with 'weighted' averages . Never heard it as such before but I believe it is this:

First year:

(73+70+69+67+0.5(66+56+56+41) )/(4+0.5(4))

Second year:

(72+65+62+53+53+50+0.5(40+40))/(6+0.5(2))

I don't have a calculator ATM but this should be correct.

The general form is

(sum of (score*weightage)/(sum of (weightage*number of scores))

EDITED So sorry I believe this is correct now.

Edit number 2: your score is 64.75,56.43. URL is here http://www.wolframalpha.com/input/?i=evaluate+%2873%2B70%2B69%2B67%2B0.5%2866%2B56%2B56%2B41%29+%29%2F%284%2B0.5%284%29%29++%2C++%2872%2B65%2B62%2B53%2B53%2B50%2B0.5%2840%2B40%29%29%2F%286%2B0.5%282%29%29

FYI, wolframalpha is a good website to do calculations

4 points

·
19th Jun 2011

stuffonstuff mentioned the easier way to do this (partial fractions), but a trig substitution can work here. Let x = 2 sec t. Then dx = 2 sec t tan t dt. Plug everything in and simplify and you should get csc(t)/2, which hopefully you know how to integrate. If you convert your answer back in terms of x, it may look a little different than what you'd get with partial fractions, but after playing around with some log rules, you get the same answer.

4 points

·
2nd Nov 2015

4 points

·
14th Feb 2012

The best resource, in my opinion, is Khan Academy:

They have easy to follow videos of all kinds of levels of maths, and practice exercises. A lot of people who aren't quite understanding certain concepts find it really helpful.

4 points

·
20th Aug 2021

In decreasing importance (IMO):

- Inject the triangle inequality into your bones. Learn it for general metric spaces (which also means, learn what metric spaces are). Draw out the “triangles” involved in your triangle inequalities.
- Think about how you will take notes, study, do homework. Do all of the organizational work ahead of time. Learn LaTeX if you have time: https://www.overleaf.com/
- Peruse through your textbook, find some interesting-looking theorems, find the definitions of the terms in the theorem, look at examples to understand the definitions. Google more examples and plot weird functions with https://www.desmos.com/ .

No point in doing any deep self-studying ahead of time. That’s what the class is for. Better to figure out how to make your learning more efficient.

4 points

·
13th Apr 2011

From the physics angle, a moment is a measure of the influence a force has on the rotation of a system (ie torque). I find this relatively intuitive: if you have a force applied to an object offset from its center of mass, there is a tendency to produce rotation. The larger the offset, the larger the rotational effect (thus a long wrench can produce more torque than a short one with the same applied force). The mathematical concept of moments seems to be related, but is clearly a different beast. I feel like it's a generalization of the physics usage, but I don't quite see how it is (yet).

The only etymology that I could really see is that the term somehow is derived from momentum, which is itself derived from the Latin terms meaning movement and moving power.

4 points

·
4th Aug 2020

It's an unsolvable integral. If you solve this as a definite integral (0 to infinity) using complex analysis or fourier transforms, the answer is pi/2.

Here is a link to the solution where complex integration is done:

https://socratic.org/questions/integration-of-sinx-x-from-0-to-infinity

4 points

·
26th Mar 2015

I also highly suggest Symbolab it is free and there is an app if you want it on the phone. Shows step by step how to solve. WolframAlpha is awesome and powerful, but can't beat the free step by step on a web browser.

4 points

·
2nd Jun 2017

Delta (𝛿) is just used to measure a distance away from some x value.

And we use epsilon (ε) to measure a distance away from the value of the limit L.

Let's say the limit of some function is 7, as x approaches 1: lim (x->1) f(x) = 7.

Then we might ask: "what values of x make the value of the function fall within the interval (7 - ε, 7 + ε)?"

The answer is some interval: (1 - 𝛿, 1 + 𝛿).

EDIT: Here is an example. This graph shows the function f(x) = 2x+5. The limit of this function as x approaches 1 is 7:

lim (x->1) 2x+5 = 7

But how can we **prove** that? Well, how close do want the function to be to 7? For any value you choose for how close to 7 you want to be (that's ε), I can find an interval that x can be in to meet that target (that's 𝛿).

Right now on the graph I have ε set to 0.5. So in order for f(x) to be within 0.5 of 7 (in the interval (7-0.5, 7+0.5)), x has to be within a 𝛿 of 0.25 (in the interval (1-0.25, 1+0.25)).

If you now change ε, you'll have to change 𝛿.

4 points

·
3rd Oct 2016

Sketch sketch sketch sketch sketch!

Try sketching y = (x+1)(x-2)(x+3) and seeing where it's above the x-axis.

You've got a cubic graph that crosses the x-axis at x =-3, x=-1 and x =2, and the y-axis at -6. It's a positive cubic, so it starts in the bottom left ( y < 0 if x < -3 ). Between -3 and -1, it becomes positive; it's negative between -1 and 2, then positive again for x > 2.

4 points

·
31st Jul 2018

This list is extremely useful, thanks!

Check also fxSolver as a candidate for the "Graphing & Visualizing Mathematics" category and as a lighter, free alternative to Wolfram Alpha.

4 points

·
25th Sep 2020

In general, for a set of n items there are n! ways to rearrange it. This set has the special fact that there are duplicates, which reduces that number by a lot. here is a brilliant.org article that explains your problem.

4 points

·
27th Aug 2018

It would be helpful if you elaborated on what kind of math you currently do. Mathematics is also a very broad field so its difficult to give specific advice for understanding the whole of mathematics better

That being said, if you are trying to better understand mathematics, i would recommend just trying to read stuff about mathematics wherever you want and when you get to a point where you don't understand something, look it up. Wikipedia, Wolfram Mathworld, brilliant.org, and The Art of Problem Solving are all good resources for looking stuff up when you get confused.

You might also try learning more about proofs if you haven't already done so because proving theorems is really what pure mathematics is about. Unfortunately i don't know any great resources on learning to prove stuff, specifically, but math textbooks written for math majors will almost certainly include proofs as a major part as opposed to focusing on applications and computation.

Finally, if you ever find yourself using math for whatever you are doing professionally, try and learn more about the theory behind what you are doing

Hope that helps Tl;dr read about math a lot

4 points

·
1st Nov 2017

Wikipedia page on factorial discusses gamma function and pi function as ways of extending the idea of factorials to non-integer arguments. The pi function for n does yield n! for integer values of n.

Are you maybe doing binomial expansion with a fractional power? That does not require factorial of nonintegers.

https://brilliant.org/wiki/fractional-binomial-theorem/

Edit: Part b in your problem is precisely this. Read the link.

4 points

·
25th Feb 2011

Your formula is right

y = x - 0.0275x.

You can factor out the x on the right hand side to get

y = x(1 - 0.0275) = 0.9725x

Divide both sides by 0.9725 to get

x = y/0.9725.

Plug in the amount you want to get into y. This will give you the amount you should list the thing at.

x = 52/0.9725 ≈ 53.47.

Alternatively, use wolframalpha.

4 points

·
10th Aug 2011

I found this comment on a khan academy video when I was learning rate of change:

> It's not a coincidence. Think of the definition of the derivative, lim Δr->0 [ΔA/Δr]. The difference in area of two circles is the area of the bigger one minus the smaller one, i.e, a circular ring. The difference in radius of two circles is the width of the ring. As Δr->0, you are calculating the area of a circular ring as the ring gets thinner and thinner. When you finish taking the limit, you have taken the area of an infinitely thin ring, which is the circumference.

> The same thing happens with volume of a sphere, 4πr³/3. dV/dr=4πr². The same reasoning, you are finding the volume of the difference of two spheres, i.e. a hollow spherical shell. As Δr->0 you are finding the volume of an infinitely thin sphere, i.e., the surface area. I spent a long time thinking about this and it's really cool, I know "infinitely small" is informal but it helps a lot in visualizing things.

- Gershom Joshua Haber, here

4 points

·
22nd Mar 2011

Just in case you didn't know about it, Khan Academy is a useful resource.

But let's see what others have to say about how to tackle her problems, because I don't know what it takes to help somebody with these problems and whether videos a la Khan Academy are sufficient for that.

4 points

·
1st Jul 2021

Hi. I really enjoy this book. I say in present tense as I often go back: "Mathematics: From the Birth of Numbers" by Jan Gullberg. The author wrote the book for the benefit of his son who was entering an academic program. He wanted something that encompassed math technique and history. It is a brilliant work. It wont be your only stop but it is a great map of the concepts up to and lightly into calculus. This history may help you connect to the technical aspects. https://www.amazon.com/Mathematics-Birth-Numbers-Jan-Gullberg/dp/039304002X.

Best of luck.

4 points

·
8th Oct 2020

I'm 33 and I did not learn coding in school either.

<em>How to Prove It: A Structured Approach, 2nd Edition</em> by Daniel Velleman, is *very* good. He does reference computer coding, but no background in coding or proofing is required. Literally anyone can pick it up and get started.

3 points

·
7th Sep 2013

Have you tried online tools like Khan Academy?

The key (for me at least), is not to worry about memorizing anything. Just keep doing problems until it becomes intuition. Lots and lots of problems.

3 points

·
8th Aug 2013

Ah, yes. I see what you're doing. Yes you can rewrite it but it's incredibly messy. Here's a video from Khan Academy which rewrites an easier equation in standard from. They take a = - f, b = 0, c = f, and d = 0. You can extend their method to your more general hyperbola but it will be a lot messier.

3 points

·
7th Mar 2013

I would highly suggest checking out Khan Academy. I'm in the process of getting my undergrad after being out of school for 10+ years and this website has been invaluable to me.

3 points

·
4th Jan 2013

I'd recommend going to the Khan Academy's website and brushing up on calculus. I'm currently using them to learn the math I neglected to learn in my youth.

Their address is: http://www.khanacademy.org

3 points

·
9th Nov 2012

check out lattice multiplication, it's pretty awesome. Here's the Khan Academy video for it: http://www.khanacademy.org/math/arithmetic/multiplication-division/v/lattice-multiplication If you want to learn how to do it the old fashioned way, there's a video on Khan Academy for that, too. My students love to review their math concepts w/Khan Academy.

3 points

·
6th Dec 2011

Since yesterday you can even use google to do this

3 points

·
14th May 2019

https://www.desmos.com/calculator/3hhnsokcfu

So you've got a sort of parabolic hollow cone shape when rotating around the x axis.

The easiest integral is probably over x from 0 to 2 of the outer circle - pi(4-x^(2))^2 minus the inner circle - pi(4-2x)^2 . This gives a polynomial integral which should be easy to evaluate once multiplied out.

3 points

·
4th May 2019

Yep. Try graphing each function separately.

f(x)=3x from -1 to 0 is, well, going from -3 to 0.

Note that if you are actually trying to find the area you have to be careful when breaking things up like this. The area is actually the integral of the absolute value, which you can't find very easily. It works here because the first part is always negative, the second part is always positive, and the final result is always positive. But if you're integrating over some parts positive and others negative, the result is not the same as the area or the negated area.

3 points

·
26th Apr 2019

For the calculator thing - it'll be in degrees mode!

So I think you're getting confused by notation that you probably haven't learnt yet - that U is actually called the union, and "A U B" is a set which has everything which is in A, and everything that is in B. So the range actually includes both sides - which makes sense if you look at the graph (and this is where the x/y value thing comes in handy). You can get any value above 1, and any value below -1 - but try to get a value that's in between, and you're not gonna get any luck.

The allowed x coordinates is called the domain, and although you can specify it, most functions have most of the real numbers as a domain, except numbers where it makes no sense. For example, 1/x has all the real numbers as a domain, except 0, because as you might know 1/0 doesn't really have a value. Using this sort of rule of thumb, can you figure out the domain of sec? ANSWER (click): >!It's all the real numbers that are not an integer multiple of pi - as cos (npi), where n is an integer, is 0, and then you have 1/cos(npi) being a 1/0 situation, and so we cut it out of the function so we live with no issues!<

3 points

·
17th Apr 2019

That's incorrect. sqrt() is a function, defined as the primary branch of the inverse of the square function.

There's no point getting hung up on it; it's just a definition. We want sqrt to be a function because functions are nice. If it were a multi-function then it would have different properties.

3 points

·
23rd Jul 2011

> I know about how the constant cancels out, but that's not the issue I'm having with this. :)

Okay, sorry if I misunderstood.

> If one solution is true for all x (I suppose it is?) and the other for a small, restricted range, and W|A gives the second choice by default - with no mention of it being restricted - isn't that a bug?

You have a valid complaint, but I wouldn't call it a bug. The trig identity does not restrict the domain, but rather the simplification log(x^2 ) = 2 log(x), which then restricts the values to positive x. Wolfram alpha makes this simplification in general with no mention of the domain: see here. I wouldn't call that a bug, but rather a choice that they make to try to make alpha's output better. (Mathematica 8 does not make this simplification by the way, unless you tell it to assume x>0 or something similar.)

If I was doing things by hand, I would probably put in an absolute value on the argument of log when I do the antiderivative. In this case everything should go through fine, essentially because you are restricting the domain in the first step. Alpha doesn't do this: see here. The reason I assume is that alpha likes to work with complex numbers, which makes log(x) a perfectly good antiderivative, even for negative x.

3 points

·
17th Jul 2011

How about the complex zeros of a function, Euler's formula, complex conjugates, the cube root of -8, quantum mechanics? See other applications of complex numbers.

Aren't complex numbers, although perhaps interpreted otherwise graphically, considered scalar quantities? Could you not have a 2-tuple vector with complex numbers, like [2+i,3-2i]? Or could you again translate these imaginary parts into additional dimensions, like [2, 3, 1, -2] or [2, 3, -1] or [2, 3, -3] or [2, 3, 3]?

I haven't taken a complex analysis course yet, though.

3 points

·
11th Jul 2011

sin 1 = .8415... Look at the "triangle" section. Sin would be the y part and cos would be the x part. Just figure out the length of the legs of the triangle.

For tangent: Sin/Cos = Tan